$\overline{AB}$ = $\sqrt{65}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $\sqrt{65}$ $?$ $ \sin( \angle ABC ) = \frac{7\sqrt{65} }{65}, \cos( \angle ABC ) = \frac{4\sqrt{65} }{65}, \tan( \angle ABC ) = \dfrac{7}{4}$
$\overline{AB}$ is the hypotenuse $\overline{AC}$ is opposite to $\angle ABC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle ABC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{\sqrt{65}} $ $ \overline{AC}=\sqrt{65} \cdot \sin( \angle ABC ) = \sqrt{65} \cdot \frac{7\sqrt{65} }{65} = 7$